2011年大连ACM网络赛 hdu 4002 Find the maximum

Find the maximum

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)
Total Submission(s): 276    Accepted Submission(s): 141

Problem Description

Euler's Totient function, φ (n) [sometimes called the phi function], is used to determine the number of numbers less than n which are relatively prime to n . For example, as 1, 2, 4, 5, 7, and 8, are all less than nine and relatively prime to nine, φ(9)=6. 
HG is the master of X Y. One day HG wants to teachers XY something about Euler's Totient function by a mathematic game. That is HG gives a positive integer N and XY tells his master the value of 2<=n<=N for which φ(n) is a maximum. Soon HG finds that this seems a little easy for XY who is a primer of Lupus, because XY gives the right answer very fast by a small program. So HG makes some changes. For this time XY will tells him the value of 2<=n<=N for which n/φ(n) is a maximum. This time XY meets some difficult because he has no enough knowledge to solve this problem. Now he needs your help.

 

Input

There are T test cases (1<=T<=50000). For each test case, standard input contains a line with 2 ≤ n ≤ 10^100.

 

Output

For each test case there should be single line of output answering the question posed above.

 

Sample Input

2 10 100

 

Sample Output

6 30

Hint

If the maximum is achieved more than once, we might pick the smallest such n.  

 

Source

The 36th ACM/ICPC Asia Regional Dalian Site —— Online Contest

 

Recommend

lcy

 

Statistic |  Submit |  Discuss |  Note 题意：
寻找 1-n中  n/φ (n) 的最大值 其中 φ (n)为欧拉函数

分析：

我遇到这个题首先看到数据量10^100,就觉得这题可能有规律，于是写了一个c++程序打了一个表看了一下

#include 

#include 

#include 

#include 

#include 

#include 

#include 

#include 

#include 

#include 

#include 

using namespace std;

#define inf (1<<30)

__int64 a[1001001];

void all()

{

    __int64 i,j;

    for(i=1;i<=1001000;i++)

    a[i]=i;

    for(i=2;i<=1001000;i++)

    {

       if(a[i]==i)

       {

           for(j=1;j*i<=1001000;j++)

           a[i*j]=a[i*j]/i*(i-1);

       }

    }

}

ofstream fout("c:\ii.out");

int main()

{

    double M=0;

    int i,j,k;

    int n,ans;

    all();

    for(j=1;j<100;j++)

    {



        n=j;

        ans=0;

        M=0;

        for(i=0;i<=n;i++)

        {

            if(M<((double)i/(double)a[i]))

            {

                M=((double)i/(double)a[i]);

                ans=i;

            }

        }

        cout<

就发现

1               1

2-5            2=1*2

6-29          6=1*2*3

30-209      30=1*2*3*5

210-2309   210=1*2*3*5*7

.....

原来就是一个质数相乘

然后就用java打了表 过了！

现在在网上找了一个证明，挺好的。。供有行人学习

证明链接： blog.csdn.net/tanhaiyuan/article/details/6747408

import java.math.BigInteger;
import java.util.Scanner;


public class Main
{
    static int pcount=0;
    static int get[]= new int[1000];
    public static  void getprime()
    {
        boolean sum[]= new boolean[1000001];
        int i,j;
        for(i=2; i<100000; i++)
        {
            if(i%2==1)sum[i]=true;
        }
        int temp=(int)Math.sqrt(100000);
        for(i=3; i0)
        {
            t--;
            n=s.nextBigInteger();
            if(n.compareTo(b6)<0)
            {

                if(n.compareTo(BigInteger.ONE)==0)
                {
                    System.out.println("1");
                }
                else System.out.println("2");
            }
            else
                for(i=1; i<102; i++)
                {
                    if(n.compareTo(val[i])<=0)
                    {
                        System.out.println(val[i-1]);
                        break;
                    }
                }
        }

    }
}